Mechanical gears are used in sets to multiply torque (rotational force) or speed (rotational velocity). Torque and speed are inversely related in gear sets; so as torque goes up, speed goes down proportionately. One gear driving another for these purposes is a gear set, systems of gear sets are called transmissions or gear boxes, and series of either of these are termed gear trains or drive lines. Calculating the overall geared torque through drive lines is straightforward and systematic.
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Define the geared drive line. Suppose you want to purchase a tractor to pull heavy loads. It is already determined that the tractor needs to apply at least 2268 kg (5,000 pounds) of pulling force to the ground to haul the loads. The tractor's engine has a maximum of 125-30cm-0.5 kg (125-foot-lb). of torque at 2,500 revolutions per minute (rpm). The wheels on the tractor are 120 cm (4 feet) tall, and the rear-axle reduction gear ratio is 10-to-1. With these parameters clearly defined, you can calculate the required gear ratio to provide the 2268kg. of pulling force.
Calculate the required torque at the tractor wheel drive axle. Because the wheels are 120 cm (4 feet) tall, their radius is 60 cm (2 feet). The axle shaft torque required to provide 2268kg (5,000 pounds). of force at a radius of 60 cm (2 feet) from the axle is 2268 x 60 cm =136080-30 cm-kg (5000 x 2 feet = 10,000-foot-lb). of geared torque needed at the axle. *For the purposes of simplifying this article the 10,000-foot-lb example shall be used.
Figure the required transmission-to-rear axle (drive shaft) output torque. The transmission output drives the axle shaft through a 10-to-1 bevel gear set, which changes the direction of the torque (fore-aft to side-to-side) while multiplying the torque by a factor of 10. If 10,000-foot-lb. of torque is required at the axle, the transmission would have to supply 10,000-foot-lb./10-to-1 ratio, or 1,000-foot-lb. of geared torque to the drive shaft.
Calculate the required low gear ratio in the transmission to provide 1,000-foot-lb. of torque to the drive shaft. Because the engine produces 125-foot-lb. of torque to the transmission input through the clutch, divide the output torque by the input torque to indicate the required transmission gear reduction. In this example, 1000-foot-lb./125-foot-lb. = 8. Therefore, the transmission would have to provide a total reduction of 8-to-1 to satisfy the torque multiplication requirement. With a geared torque multiplication of 8-to-1 in the transmission and 10-to-1 in the drive axle, the overall 8 x 10 = 80-to-1 geared torque multiplication would supply the required 10,000-foot-lb. of geared torque to the axle, and the 2268kg. pulling force you are looking for with this tractor.
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